Sol. We observe that the smaller of the indices of sin x and cos x in the two integrals to be connected are m and n–2 respectively. So, let P = sin m+1 x cos n–1 x

⇒ = (m + 1) sin m x cos n x –(n –1) sin m+2 x cos n–2 x

⇒ = (m + 1) sin m x cos n x –(n –1) sin m x sin 2 x cos n–2 x

⇒ = (m + 1) sin m x cos n x – (n –1)

sin m x cos n–2 x (1 – cos 2 x)

⇒ = (m + 1) sin m x cos n x – (n –1)

sin m x cos n–2 x + (n –1) sin m x cos n x

⇒ =(m + n) sin m x cos n x –(n –1) sin m x cos n–2 x

Integrating both sides with respect to x, we get

P = (m + n) sin m x cos n x dx – (n –1) sin m x cos n–2 x dx

⇒ (m + n) sin m x cos n x dx = P + (n –1) sin m x cos n–2 x dx

⇒ sin m x cos n x dx = + sin m x cos n–2 x dx